Double angles and the triangle method

What you will be able to do

• find $\sin(\cos^{-1}x)$, $\cos(\sin^{-1}x)$, $\tan(\sin^{-1}x)$, $\sin(\tan^{-1}x)$ and $\cos(\tan^{-1}x)$ from a right triangle;
• write $2\tan^{-1}x$ as one inverse sine, cosine or tangent, and say where each form is true;
• find the trigonometric function of a sum of inverse functions.

The triangle method

An inverse function is an angle, so a right triangle shows it. Say $\theta = \tan^{-1}x$ with $x > 0$. Then $\tan\theta = \dfrac{x}{1}$. Draw a triangle with the side opposite $\theta$ equal to $x$ and the side next to $\theta$ equal to $1$. By Pythagoras the hypotenuse is $\sqrt{1 + x^2}$. Now read off

In the same way, with the hypotenuse $1$:

• $\theta = \sin^{-1}x$: the opposite side is $x$ and the adjacent side is $\sqrt{1 - x^2}$. So $\cos(\sin^{-1}x) = \sqrt{1 - x^2}$ and $\tan(\sin^{-1}x) = \dfrac{x}{\sqrt{1 - x^2}}$.
• $\theta = \cos^{-1}x$: the adjacent side is $x$ and the opposite side is $\sqrt{1 - x^2}$. So $\sin(\cos^{-1}x) = \sqrt{1 - x^2}$.

The square root is the positive one, and this is why. The range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, where the cosine is not negative. The range of $\cos^{-1}$ is $[0, \pi]$, where the sine is not negative. These formulas stay true for a negative $x$: the triangle gives the sizes, and the branch gives the sign.

Double angles

Let $x = \tan t$, so that $t = \tan^{-1}x$ lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. Then $2\tan^{-1}x = 2t$ lies in $(-\pi, \pi)$. The double angle formulas give

To write $2t$ as the inverse of one of these, the angle $2t$ must lie in the branch of that inverse function.

• Inverse sine: $2t$ must be in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, which means $|x| \le 1$. Then $2\tan^{-1}x = \sin^{-1}\dfrac{2x}{1 + x^2}$.
• Inverse cosine: $2t$ must be in $[0, \pi]$, which means $x \ge 0$. Then $2\tan^{-1}x = \cos^{-1}\dfrac{1 - x^2}{1 + x^2}$.
• Inverse tangent: $2t$ must be in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, which means $|x| < 1$. Then $2\tan^{-1}x = \tan^{-1}\dfrac{2x}{1 - x^2}$.

For example, $2\tan^{-1}\dfrac{1}{3} = \sin^{-1}\dfrac{3}{5} = \cos^{-1}\dfrac{4}{5} = \tan^{-1}\dfrac{3}{4}$. All three are true here, because $\dfrac{1}{3}$ is between $0$ and $1$.

The picture compares $2\tan^{-1}x$ with each form. A form is true only inside the shaded part.

There is a similar rule for the inverse sine: $2\sin^{-1}x = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)$ for $|x| \le \dfrac{1}{\sqrt{2}}$.

A worked example

Find (a) $\sin\left(\cos^{-1}\dfrac{5}{13}\right)$, (b) $\cos\left(\sin^{-1}\dfrac{4}{5} + \cos^{-1}\dfrac{5}{13}\right)$, (c) $2\tan^{-1}\dfrac{1}{2}$ as one inverse tangent.

1. (a) The adjacent side is $5$ and the hypotenuse is $13$, so the opposite side is $12$. The sine is $\dfrac{12}{13}$.
2. (b) Let $\alpha = \sin^{-1}\dfrac{4}{5}$ and $\beta = \cos^{-1}\dfrac{5}{13}$. Then $\sin\alpha = \dfrac{4}{5}$, $\cos\alpha = \dfrac{3}{5}$, $\cos\beta = \dfrac{5}{13}$ and $\sin\beta = \dfrac{12}{13}$. So $\cos(\alpha + \beta) = \dfrac{3}{5} \times \dfrac{5}{13} - \dfrac{4}{5} \times \dfrac{12}{13} = \dfrac{15 - 48}{65} = -\dfrac{33}{65}$.
3. (c) $x = \dfrac{1}{2}$ has $|x| < 1$, so $2\tan^{-1}\dfrac{1}{2} = \tan^{-1}\dfrac{1}{1 - \frac{1}{4}} = \tan^{-1}\dfrac{4}{3}$.

Practise with new numbers

Try it in Python

The program checks the three forms of $2\tan^{-1}x$ for five numbers. For some numbers a form gives the wrong angle.

Common mistakes

• Forgetting the interval of a double angle form. $2\tan^{-1}x = \sin^{-1}\dfrac{2x}{1 + x^2}$ fails for $|x| > 1$.
• A negative root. $\sin(\cos^{-1}x)$ is never negative, because $\cos^{-1}x$ is in $[0, \pi]$.
• Dropping the square root. The third side of the triangle is $\sqrt{1 - x^2}$, not $1 - x^2$.
• Mixing up the two sides for $\tan^{-1}x$. The opposite side is $x$ and the adjacent side is $1$.

Remember these

• $\sin(\cos^{-1}x) = \cos(\sin^{-1}x) = \sqrt{1 - x^2}$ and $\tan(\sin^{-1}x) = \dfrac{x}{\sqrt{1 - x^2}}$.
• $\sin(\tan^{-1}x) = \dfrac{x}{\sqrt{1 + x^2}}$ and $\cos(\tan^{-1}x) = \dfrac{1}{\sqrt{1 + x^2}}$.
• $2\tan^{-1}x$ is $\sin^{-1}\dfrac{2x}{1 + x^2}$ for $|x| \le 1$, $\cos^{-1}\dfrac{1 - x^2}{1 + x^2}$ for $x \ge 0$, and $\tan^{-1}\dfrac{2x}{1 - x^2}$ for $|x| < 1$.
• For a sum of two inverse functions, find the sine and cosine of each, then use the addition formula.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\sin\left(\cos^{-1}\dfrac{7}{25}\right)$.

2. Find $\tan\left(\sin^{-1}\dfrac{3}{5}\right)$.

3. Find $\cos\left(\tan^{-1}2\right)$.

4. Write $2\tan^{-1}\dfrac{1}{4}$ as one inverse tangent.

5. For which $x$ is $2\tan^{-1}x = \sin^{-1}\dfrac{2x}{1 + x^2}$ true?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions.
• OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions.
• OpenStax, Calculus Volume 1, Section 1.4: Inverse Functions.